/*
	author: TangQiao , Wind @ Beijing Normal University

	problem name: Error Correction
	
	source :  BNU Online Judge
	
	problem type: 模拟题
	
	problem description: 找出奇偶校验码中的错误并试着修正. 
	
	problem solution: 计算每行和列的1的个数,然后看有多少出错的行和列.
	如果行列各一个则可以修正,大于2个则不能.没有则正确.

	
	date : 2005.7.16 北师大个人练习赛
	
*/
#include <stdio.h>
#include <string.h>
#include <math.h>

int shu[100][100];
int n;
int ni,nj;

int check (int &ni, int &nj)
{
	int i,sum1,sum2;
	for (i=1,sum1=0;i<=n;i++)
		if (shu[i][0]%2==1) 
		{
			sum1++;
			ni=i;
		}
	if (sum1>=2) return -1;
	for (i=1,sum2=0;i<=n;i++)
		if (shu[0][i]%2==1)
		{
			sum2++;
			nj=i;
		}
	if (sum2>=2) return -1;
	if (sum1==0 && sum2==0) return 0;
	return 1;
}

main()
{
	int i,j;
	while (1)
	{
		scanf("%d", &n);
		if (n==0) break;
		memset(shu,0,sizeof(shu));
		for (i=1;i<=n;i++)		
			for(j=1;j<=n;j++)
			{
				scanf("%d", &shu[i][j]);
				shu[i][0]+=shu[i][j];
				shu[0][j]+=shu[i][j];
			}
		j=check(ni, nj);	
		if (j==0) printf("OK\n");
		else if (j==-1) printf("Corrupt\n");
		else printf("Change bit (%d,%d)\n",ni,nj);

		
	}
	return 0;
}